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Velocity and acceleration analysis analysis using vectors Problem 1

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Title: Velocity and acceleration analysis analysis using vectors Problem 1


1
Velocity and acceleration analysisanalysis using
vectorsProblem 1 
Link 2 rotates at a constant angular velocity.
Determine the velocity and acceleration of points
C and D.
 
2
Solution 1 Velocity analysis for C
VC4VC3VC3A2VA2
VA2 1 rad/s2in 2in/s Normal to link 2
Hence direction and magnitude are both known for
VA2
VC3A2 is normal to link 3. Only direction is
known
VC4VC3 is horizontal. Only direction is known
Absolute velocity vectors VA2 and VC3 start from
Ov. Relative velocity vectors never start from
Ov.
A2
VA2 2in/s D, M
VC3A2 1.45 in/s (approx) D, M?
Ov
C3
VC4VC3 1in/s (approx) D, M?
VC3A2 1.45 in/s Hence w3 VD3A2
/AC 1.45/80.18 rad/s
Scale 1 in1 in/s
3
Solution 1 Acceleration analysis for C
aC4aC3aC3A2aA2anC3A2 atC3A2anA2atA2
atC3A2 Only Direction is known normal
to AC
aC4aC3 Direction is known - horizontal
anC3A2 w23 AC 0.1828 in/s2 0.26 in/s2
Direction- Radially inward along AC
aA2 w22 OA (1 rad/s)2 2 in2in/s2 Direction
radially inward along O2A
aC4aC3 1.48 in/s (approx) D, M?
Oa
C
The vector diagram
aA2 w22 OA (1 rad/s)2 2 in 2in/s2 D, M
atC3A2 1.38 in/s2 (approx) D, M?
atC3A2 1.38 in/s2 Hence a3 atC3A2
/AC 1.38/80.17 rad/s2
anC3A2 w23 AC 0.1828 in/s2 0.26 in/s2
(approx) D, M
4
Solution 1 Velocity analysis for D
VB5VB3VB3A2VA2 VB5D6VD6
VB3A2 w3AB VD3A2 (AB/AC) 0.18 rad/s4
in 0.72 in/s Normal to link 6.
VA2 1 rad/s2in 2in/s Normal to link 2
VD6 is normal to link 6. Only direction is known
VB5D6 is normal to link 5. Only direction is
known
Absolute velocity vectors VA2 and VC3 start from
Ov. Relative velocity vectors never start from
Ov.
VB5D6 1.3 in/s (approx) D, M?
VB3A2 0.72 in/s (approx) D, M
A
D
VD6 1.5 in/s Hence w5 VD6 /O6D 1.5/40.38
rad/s
VD6 1.5 in/s D, M?
VB5D6 1.3 in/s Hence w5 VB5D6 /BD 1.3/40.33
rad/s
VA2 2in/s D, M
B
Ov
Scale 1 in1 in/s
5
Solution 1 Acceleration analysis for D
aB5aB3aB3A2aA2 aB5D6aD6 anB5D6
atB5D6anD6atD6anB3A2 atB3A2anA2atA2
atD6 Only Direction is known normal to
O6D
anD6 w26 O6D (0.38 rad/s)2 4 in0.57
in/s2 Direction radially inward along O6D
anB5D6 w52BD 0.33240.44 in/s2 Direction
radially inward along BD
atB3A2 Only Direction is known normal
to AC
atB5D6 a3AB atC3A2(AB/AC) 1.384/80.69
in/s2 Direction normal to BD
anB3A2 w23 AB 0.1824 in/s2 0.13 in/s2
Direction- Radially inward along AC
aA2 w22 OA (1 rad/s)2 2 in2in/s2 Direction
radially inward along O2A
6
Solution 1 Acceleration analysis for D
aB5aB3aB3A2aA2 aB5D6aD6 anD6atD6anB5D6
atB5D6 anA2atA2 anB3A2 atB3A2
Since D cannot be located using the current
order of vector addition, locating B is also
difficult.
Where is B3 ?
Where is D6? Where does anB5D6 begin?
Oa
Scale 1 in 0.5 in/s2
7
Solution 1 Acceleration analysis for D
Oa
Change order of addition of vectors
atD6 4.1 in/s2 D, M ?
aB5aB3aB3A2aA2 aB5D6aD6 anD6atD6anB5D6
atB5D6 anA2 anB3A2 atB3A2atA2
aB5aB3aB3A2aA2 aB5D6aD6 anD6atD6anB5D6
atB5D6 anA2atA2 anB3A2 atB3A2
atB3A2 2.8 in/s2 D, M?
aD 4.05 in/s2
Scale 1 in 0.5 in/s2
B
D
8
Velocity and acceleration analysisanalysis using
vectorsProblem 2
    Determine the linear acceleration of AP4 if
a20
9
Solution 2 Acceleration analysis for P
w2 Sense of rotation is found from direction of
VP2
Analyze available information
  •  VP2 10.0 in/sec. OPP0.95.
  • Hence w2 VP2/OPP 10.53 rad/sec.
  • Hence anP2 w22 OPP105.34 in/sec2
  • a2 0.
  • Hence atP2 0
  • VP4P2 10.77 in/sec.
  • Hence acP4P2 w2VP4P2 10.53 10.77113.38
    in/sec2.

Vector equation
aQ4aP4aP3 aP4P2aP2asP4P2 acP4P2anP2atP2
10
Solution 2 Acceleration analysis for P
aP4 D, M?
asP4P2 D, M?
asP4P2 55
aP4 18
aP2anP2 (Since atP20) w22 OPP 105.34 D, M
acP4P2 2w2VP4P2 113.38 in/s2
aP4 Direction Along OP (OpPO)
acP4P2 2w2VP4P2 113.38 in/s2
asP4P2 Direction -Along AP
atP2 0
anP2 w22 OPP105.34 Radially inwards along OpP
(OpPO)
11
Velocity and acceleration analysisanalysis using
vectorsProblem 3
Link 2 rotates at a constant w2 1 rad/sec. q2
60o. Find VA4 and aA4 if O2A 2.4 cm
12
Solution 3 Velocity analysis for A
VA3A4 is along the slider in 4. Only direction
is known
VA3 w2O2A 1 rad/s2.4cm 2.4 cm/s Normal to
O2A
VB4 is along the slider in 1. Only direction is
known
VA4 VB4 (since 4 is translating only, all
points on 4 have same linear velocity)
VA3 w2O2A 1 rad/s2.4cm 2.4 cm/s
A3
VA3A4 1.2 cm/s D, M?
A4
Ov
VA42.08 cm/s D, M?
13
Solution 3 Acceleration analysis for A
aA2 aA3 anA2atA2 aA4asA3A4 acA3A4
Here we observe a link 3 which slides relative
to a moving link 4. Hence there should be a
Coriolis term
atA2 a2O2A 02.40 Hence aA2 anA2
aCorioliswcontaining linkVrelcontained link
aA4 aB4 ? Along O2B4
aCA3A4 2w4 VA3A4
But w4 0 as link 4 translates relative to the
ground. Hence acA3A4 0
asA4A3 ? Along A2B4
aA2 anA2 w22O2A 2.4 cm/s2 Radially inward
from A to O2
aA4 1.2 cm/sec2 D, M ?
Oa
A4
asA3A4 2.08 cm/s2 D, M ?
aA3 aA2 anA2 2.4 cm/sec2
A3
14
Problem 4
w2 1.000
1.0
1.631
3.56
4.41
Find the relative acceleration of the slider
block with respect to the curved slot in the
mechanism shown in figure. Assume w21 unit
angular velocity clockwise, a20, O2A1 unit
length. Use graphical method of acceleration
analysis using vectors.
15
Velocity diagram
tangent
OV
VA4
0.835 w40.5
A4
VA21
VA4 A3
1.016
A2 , A3
16
Acceleration diagram
OA
w42O4A0.41
a4O4A
A4
A2 , A3
2w4VA4A3 V2A4A3/4.41/
aA4A3,sliding
tangent
normal
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